 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Diophantine And Almost Fermat (Posted on 2007-02-21) Consider three positive integers x< y< z in arithmetic sequence.

Determine analytically all possible solutions of each of the following equations:

(I) x3 + z3 = y3 + 10yz

(II) x3 + y3 = z3 - 2, with y< 116.

 See The Solution Submitted by K Sengupta Rating: 4.0000 (1 votes) Comments: ( Back to comment list | You must be logged in to post comments.) possible solutions | Comment 2 of 4 | I didn't get the meaning of "arithmetic sequence" at first.  Thanks for the additional input.

I was unable to find a solution that fit both equations.  For the first equation, (1,4,7) and (3,6,9) fit.  For the second equation, (5,6,7) works.

For a solution that fits both equations, I had first tried by combining the equations to get [x�=5yz-1].  From this, I determined that x must end in either 4 or 9 since 5yz ends in either 0 or 5.  Subtracting the two equations yields [z�-y�=5yz+1].  Using a for the arithmatic sequence step leads to [3ax�+9a�x+7a�=5x�+3ax+2a�].  However, I could not find an integer solution that satisfied both equations.
 Posted by hoodat on 2007-02-26 18:20:24 Please log in:
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